RE: [math-fun] 7.11 puzzle
At 12:51 PM 12/5/03, Rich wrote:
Probably still legal tender for small debts, but I doubt 7-11 would take them. But it raises an interesting generalization: Prices are frequently "2 for 1.29". Suppose we ignore the custom of rounding up (.645 -> .65), and allow (positive) rationals in the solution. It's not clear to me that the set of solutions is still finite, and my enumeration methods would only work if the denominators are bounded.
-----Original Message----- From: R. William Gosper To: math-fun@mailman.xmission.com Sent: 12/4/2003 11:19 PM Subject: Re: [math-fun] 7.11 puzzle
Are half-cents still legal tender?-) 1.125, 1.28, 3.125, 1.58
There are indeed plenty of rational solutions. Taking all values in cents, we're solving a + b + c + d = 711, abcd = 711000000. Using a known solution, namely the intended (316,150,125,120), as a starting point, fix a = 316 and treat b as a parameter. Then c and d are given by c + d = 395 - b, cd = 2250000/b, and so they are the roots of a quadratic with discriminant D = sqrt((395 - b)^2 - 9000000/b). So a rational solution corresponds to a rational point on the cubic curve b^3 - 790*b^2 + 156025*b - 9000000 = b*D^2. We already know rational points with b = 120,125,150, so we can find more by the chord-and-tangent method. The drawback is that the denominators get large fast. The first two all-positive solutions I find are (316, 1922/13, 4056/31, 46875/403) and (316, 956092840/8367257, 2961126750/20587579, 1329939075/9714443). P.S. There's also a very nice non-positive solution in whole cents: (316,900,-500,-5). -- Fred W. Helenius <fredh@ix.netcom.com>
FredH>There are indeed plenty of rational solutions. [...] I never imagined the chord-and-tangent method would work given a crossterm, particularly since there are four real branches instead of the usual one or two. In fact, scaling by 120 sqrt(3) and shifting by 197/100, one gets y^2 = (50 x -197)^2/7500-790/(100 x + 197) which, miraculously, has 120 degree symmetry (as well as up-down). A "sesquihyperbola" surrounding a rounded triangle. And sure enough, no line hits the curve more than three times. Rescaling to whole dollars, the addition and doubling formulas are [a,b0,c0,d0]+[a,b1,c1,d1] = [a,b2,c2,d2] and 2*[a,b,c,d] = [a,b3,c3,d3], with b2=((100*a*(b1-b0)^2*d0*d1)/(100*a*b1*d0*d1+100*a*b0*d0*d1+100*a*b1^2*d1 +100*a^2*b1*d1-711*a*b1*d1+100*a*b0^2*d0+100*a^2*b0*d0-711*a*b0*d0+1422)) 2 b2 = 100 a (b1 - b0) d0 d1/(100 a b1 d0 d1 + 100 a b0 d0 d1 2 2 2 + 100 a b1 d1 + 100 a b1 d1 - 711 a b1 d1 + 100 a b0 d0 2 + 100 a b0 d0 - 711 a b0 d0 + 1422) b3=-((100*a*b^2*(d-c)^2)/(200*a*b^3+100*a^2*b^2-711*a*b^2+711)) 2 2 100 a b (d - c) b3 = - ------------------------------------- 3 2 2 2 200 a b + 100 a b - 711 a b + 711 and cyclically, b <- c <- d <- b for the other coordinates. Assorted solutions: [49/25,2883/1820,169/62,2370/2821],[49/25,4099683/2972866,158623231/171679340,33012030/11595311],[25/26,474/325,169/100,3],[79/105,49/25,25/12,81/35],[49/25,-1805/812,5046/665,-237/1102],[79/25,-88209/50680,32761/5544,-3920/17919],[79/25,-1/20,-5,9],[5/4,-711/8134,-6889/1225,4802/415],[3/2,9604/191,-79/37436,-109443/2450],[79/25,32761/5544,-88209/50680,-3920/17919],[6/5,-6889/955,109443/8300,-1975/31706],[25/8,361/40,-474/95,-24/475],[25/8,-96/25,79/10,-3/40],[25/8,-4050/2821,-75919/273000,66248/11625],[25/8,-4050/2821,66248/11625,-75919/273000],[6/5,-27,395/12,-1/150],[6/5,-147/1300,1975/182,-169/35],[6/5,1975/182,-169/35,-147/1300],[6/5,109443/8300,-6889/955,-1975/31706],[6/5,-881877/257632,1204352/126575,-34445/189344],[49/25,-1805/812,5046/665,-237/1102],[49/25,-64/13715,44521/1456,-600795/23632] some of which came from search results: [1250/13,1896/13,169,300]/100, [1580/21,196,625/3,1620/7]/100, plus the several earlier mentioned. --rwg
participants (2)
-
Fred W. Helenius -
R. William Gosper